= Theorem =
${d\over dx}(\tanh^{-1}x)=\frac1{1-x^2}$

= Proof =
$y=\tanh^{-1}x$
$\tanh{y}=x$
$\frac{dy}{dx}=\frac1{\;\frac{dx}{dy}\;}$
 $=\frac1{\operatorname{sech}^2 y}$
 $=\frac1{1-\tanh^2 y}$
 $=\frac1{1-x^2}$

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Up: [[여러가지증명]]